Calculate the limit (1+2+...+n)/(n^2+3n+1) n --> infinite

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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lim (1+2+...+n)/(n^2 + 3n + 1)  n --> inf

By substituttion:

lim (1+2+ ..+n)/n^2 + 3n + 1) = inf/inf

But we know that:

1+ 2+ ..._+ n = n(n+1)/2

==> lim n(n+1)/2(n^2 + 3n + 1)

==> lim (n^2 + n) / 2(n^2 + 3n + 1)

Now let us multiply and divide by the heighest power n^2:

==> lim (1+ 1/n)/ 2(1+ 3/n + 1/n^2)      when n --> inf

==> lim = (1+0)/2(1+0 +0) = 1/2

Then the limit = 1/2

 

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll substitute infinite into the given expression:

 (1+2+...+inf.)/inf. = inf./inf.

As we can see, this is a case of indeterminacy

We notice that the numerator is the sum of the first n terms of an arithmetic series and the formula for the sum is:

S = (1+n)*n/2

We'll substitute the numerator by the expression of the sum:

limit [(1+n)*n/2*(n^2+3n+1)]

We'll remove the brackets and we'll draw out the constant value 1/2.

(1/2) * lim [(n^2 + n)/(n^2+3n+1)]

We notice that the numerator and denominator are polynomials that have the same degree, so the limit is the ratio of the coefficients of the terms that have the highest degree.

In this case, the ratio of the coefficients of n^2, both numerator and denominator, are 1/1.

The result of the limit is:

limit (1+2+...+n)/(n^2+3n+1) = 1/2

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To calculate lt (1+2+3+...n)/(n^2+3n+1) as n--> inf.

Solution:

We know that  the numerator ,1+2+3+....n = n(n+1)/2 .

So the given lmit becomes:

Lt {n(n+1)/2}/(n^2+3n+1} = Lt{n^2+n}/{2(n^2+3n+1}

Divide both numerator and denominator by n^2 and we get:

Lt (1+1/n)/{1+3/n+1/n^2} .

= 1/1 , as n--> 0, 1/n  and 1/n^2 approach zero.

=1

Therefore lt (1+2+3+....+n)/(n^2+3n+1) = 1 as n--> inf.

 

 

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